Tasks 03-04 - Transformations, Composition & Inverses

Section 03: Functions as Business Models

A: Skills

Problem 1: Basic Transformations (x)

Given the function \(f(x) = x^2 - 2x + 3\). Write down the function term for each transformation and simplify it.

  1. \(f(x)\) shifted up by 4 units

  2. \(f(x)\) shifted left by 3 units

  3. \(f(x)\) reflected over the \(x\)-axis

  4. \(f(x)\) stretched vertically by a factor of 2

  5. \(f(x)\) shifted right by 2 units and down by 5 units

  6. the mirror image of \(f(x)\) mirrored at the \(y\)-axis (the graph reflected over the \(y\)-axis)

  7. Write \(f\) in vertex form and give its vertex. Check your answers to a) to f): where must the vertex of each new function lie?

  1. Shifted up by 4: \(g(x) = f(x) + 4 = x^2 - 2x + 7\)

  2. Shifted left by 3, so \(x\) is replaced by \(x + 3\): \(g(x) = f(x + 3) = (x + 3)^2 - 2(x + 3) + 3 = x^2 + 6x + 9 - 2x - 6 + 3 = x^2 + 4x + 6\)

  3. Reflected over the \(x\)-axis: \(g(x) = -f(x) = -(x^2 - 2x + 3) = -x^2 + 2x - 3\)

  4. Stretched vertically by 2: \(g(x) = 2f(x) = 2x^2 - 4x + 6\)

  5. Right 2, down 5: \(g(x) = f(x - 2) - 5 = (x - 2)^2 - 2(x - 2) + 3 - 5 = x^2 - 4x + 4 - 2x + 4 - 2 = x^2 - 6x + 6\)

  6. Mirrored at the \(y\)-axis, so \(x\) is replaced by \(-x\): \(g(x) = f(-x) = (-x)^2 - 2(-x) + 3 = x^2 + 2x + 3\)

  7. Completing the square: \(f(x) = (x^2 - 2x + 1) - 1 + 3 = (x - 1)^2 + 2\), vertex \((1, 2)\). The vertex moves with the graph:

      1. up 4: \((1, 6)\). Check: \(x^2 - 2x + 7 = (x - 1)^2 + 6\) ✓
      1. left 3: \((-2, 2)\). Check: \(x^2 + 4x + 6 = (x + 2)^2 + 2\) ✓
      1. reflected over the \(x\)-axis: \((1, -2)\). Check: \(-x^2 + 2x - 3 = -(x - 1)^2 - 2\) ✓
      1. stretched by 2: \((1, 4)\). Check: \(2x^2 - 4x + 6 = 2(x - 1)^2 + 4\) ✓
      1. right 2, down 5: \((3, -3)\). Check: \(x^2 - 6x + 6 = (x - 3)^2 - 3\) ✓
      1. mirrored at the \(y\)-axis: \((-1, 2)\). Check: \(x^2 + 2x + 3 = (x + 1)^2 + 2\) ✓

Problem 2: Order of Combined Transformations (x)

  1. The graph of \(f(x) = x^2\) is mapped onto the graph of \(g(x) = 3(x + 2)^2 - 4\). Describe the transformations in the correct order and give the vertex of \(g\).

  2. The same for \(h(x) = -0.5(x - 1)^2 + 6\).

  3. Start with \(f(x) = x^2\). Give the function term if you first shift the graph up by 2 and then stretch it vertically by the factor 3. Give the function term for the other order: first stretch by 3, then shift up by 2. Compare the two vertices and explain the difference.

  4. Start with \(q(x) = x^2 - 6x + 11\). Give the function term after these steps in this order: shift right by 1, reflect over the \(x\)-axis, shift up by 3. Simplify it and check your result with the vertex of \(q\).

  5. A student describes \(k(x) = 2(x - 3)^2 + 1\) like this: “First shift the graph of \(x^2\) up by 1, then stretch it vertically by 2, then shift it left by 3.” Find the two mistakes and give the correct description.

The order for \(g(x) = a \cdot f(b(x - h)) + k\): first the horizontal stretch or compression (\(b\)), then the horizontal shift (\(h\)), then the vertical stretch or compression (\(a\), with a reflection over the \(x\)-axis if \(a < 0\)), and the vertical shift (\(k\)) last.

  1. \(g(x) = 3(x + 2)^2 - 4\):

    1. Horizontal shift by 2 to the left (\(x + 2\))
    2. Vertical stretch by the factor 3
    3. Vertical shift by 4 down

    Vertex \((-2, -4)\). Check: \(g(-2) = 3 \cdot 0 - 4 = -4\) ✓

  2. \(h(x) = -0.5(x - 1)^2 + 6\):

    1. Horizontal shift by 1 to the right (\(x - 1\))
    2. Vertical compression by the factor 0.5
    3. Reflection over the \(x\)-axis (the minus sign)
    4. Vertical shift by 6 up

    Vertex \((1, 6)\). The parabola opens downward.

  3. Up 2, then stretch by 3: \(3(x^2 + 2) = 3x^2 + 6\), vertex \((0, 6)\).

    Stretch by 3, then up 2: \(3x^2 + 2\), vertex \((0, 2)\).

    The stretch multiplies every value that is already there, so a shift made before the stretch is stretched too: the 2 becomes 6. That is why the vertical shift comes last, as in \(a \cdot f(x) + k\).

  4. Shift right by 1: \(q(x - 1) = (x - 1)^2 - 6(x - 1) + 11 = x^2 - 2x + 1 - 6x + 6 + 11 = x^2 - 8x + 18\)

    • Reflect over the \(x\)-axis: \(-x^2 + 8x - 18\)
    • Shift up by 3: \(g(x) = -x^2 + 8x - 15\)

    Check with the vertex: \(q(x) = (x - 3)^2 + 2\) has the vertex \((3, 2)\). Right by 1 gives \((4, 2)\), the reflection gives \((4, -2)\), up by 3 gives \((4, 1)\). Indeed \(g(4) = -16 + 32 - 15 = 1\), and the vertex formula gives \(x_v = -\frac{8}{2 \cdot (-1)} = 4\) ✓

  5. Mistake 1: \(x - 3\) shifts the graph to the right, not to the left (“minus shifts right, plus shifts left”).

    Mistake 2: the shift up by 1 must come after the stretch. First up 1, then stretch by 2 gives \(2(x^2 + 1) = 2x^2 + 2\), so the shift would be 2, not 1.

    Correct description: shift right by 3, stretch vertically by 2, shift up by 1. Vertex \((3, 1)\).

Problem 3: Composition Basics (x)

Given the functions

  • \(f(x) = 3x - 2\)
  • \(g(x) = x^2 + 1\)
  • \(h(x) = \sqrt{x}\)

Determine:

  1. \((f \circ g)(x)\)

  2. \((g \circ f)(x)\)

  3. \((f \circ g)(3)\)

  4. \((g \circ f)(2)\)

  5. \((h \circ g)(4)\)

  6. Give the domain of \((h \circ f)(x)\) and the domain of \((h \circ g)(x)\).

  1. \((f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 1\)

  2. \((g \circ f)(x) = g(f(x)) = g(3x - 2) = (3x - 2)^2 + 1 = 9x^2 - 12x + 5\)

    The results of a) and b) differ: the order of a composition matters.

  3. \((f \circ g)(3) = 3 \cdot 3^2 + 1 = 27 + 1 = 28\). Check: \(g(3) = 10\) and \(f(10) = 28\) ✓

  4. \((g \circ f)(2) = 9 \cdot 2^2 - 12 \cdot 2 + 5 = 36 - 24 + 5 = 17\). Check: \(f(2) = 4\) and \(g(4) = 17\) ✓

  5. \((h \circ g)(4) = h(g(4)) = h(17) = \sqrt{17} \approx 4.12\)

  6. The inner function comes first. Its values must lie in the domain of \(h\), i.e. they must not be negative.

    • \((h \circ f)(x) = \sqrt{3x - 2}\): \(f\) is defined for all \(x\), and \(3x - 2 \geq 0 \iff x \geq \frac{2}{3}\). Domain: \(\left[\frac{2}{3}; \infty\right)\)
    • \((h \circ g)(x) = \sqrt{x^2 + 1}\): a square is never negative, so \(x^2 + 1 \geq 1 > 0\) for every \(x\). Domain: \(\mathbb{R}\)

Problem 4: Inverse Functions (x)

In a) to c), determine the inverse function, give the domain and the range of the function and of its inverse, and verify your inverse function with \(f(f^{-1}(x)) = x\) (for \(g\) and \(h\) in the same way).

  1. \(f(x) = 5x - 4\)

  2. \(g(x) = \frac{x + 2}{3}\)

  3. \(h(x) = 2x^3 - 1\). Give a reason why \(h\) has an inverse function.

  4. Decide whether \(k(x) = x^2 - 8x + 19\) has an inverse function on \(\mathbb{R}\). Give a reason with two different inputs that have the same output.

  1. \(y = 5x - 4\). Swap \(x\) and \(y\): \(x = 5y - 4\). Solve for \(y\): \(y = \frac{x + 4}{5}\).
    • \(f^{-1}(x) = \frac{x + 4}{5}\)
    • Domain and range: \(\mathbb{R}\) for \(f\) and for \(f^{-1}\)
    • Verify: \(f(f^{-1}(x)) = 5 \cdot \frac{x + 4}{5} - 4 = x + 4 - 4 = x\) ✓
  2. \(y = \frac{x + 2}{3}\). Swap: \(x = \frac{y + 2}{3}\). Solve: \(3x = y + 2 \iff y = 3x - 2\).
    • \(g^{-1}(x) = 3x - 2\)
    • Domain and range: \(\mathbb{R}\) for \(g\) and for \(g^{-1}\)
    • Verify: \(g(g^{-1}(x)) = \frac{3x - 2 + 2}{3} = \frac{3x}{3} = x\) ✓
  3. Reason: \(x^3\) always increases when \(x\) increases, so \(2x^3 - 1\) is always increasing. Each output comes from exactly one input, so \(h\) is one-to-one.
    • \(y = 2x^3 - 1\). Swap: \(x = 2y^3 - 1\). Solve: \(y^3 = \frac{x + 1}{2} \iff y = \sqrt[3]{\frac{x + 1}{2}}\) (the cube root exists for every number and keeps its sign)
    • \(h^{-1}(x) = \sqrt[3]{\frac{x + 1}{2}}\)
    • Domain and range: \(\mathbb{R}\) for \(h\) and for \(h^{-1}\)
    • Verify: \(h(h^{-1}(x)) = 2 \cdot \frac{x + 1}{2} - 1 = x\) ✓
  4. No. \(k(3) = 9 - 24 + 19 = 4\) and \(k(5) = 25 - 40 + 19 = 4\): two different inputs have the same output, so \(k\) is not one-to-one. The graph is a parabola with the axis of symmetry \(x = -\frac{-8}{2 \cdot 1} = 4\), and points at the same distance from the axis have the same height. The horizontal line \(y = 4\) hits the graph twice (horizontal line test).

B: Set Up & Check

Problem 5: From Text to Functions (xx)

Translate each statement into a function. Then answer the check question: it tells you whether your translation works.

  1. “A swimming school rents a pool for its courses. Its monthly costs for \(x\) course participants are \(C(x) = 0.2x^2 + 25x + 4800\) EUR. From January on, the rent rises by 600 EUR per month.” Give the new cost function \(C_1(x)\) and name the transformation.

    Check: What are the costs in a month without participants, before and after the rent increase? Does the difference match the statement?

  2. “A garden centre sells \(D(p) = 800 - 25p\) bags of potting soil per week at a price of \(p\) EUR per bag. During a spring promotion, it sells 20 % more bags at every price.” Give the demand function \(D_1(p)\) during the promotion and name the transformation.

    Check: At 32 EUR per bag, nobody buys potting soil. Is this still true during the promotion?

  3. “A company rents out beach chairs (roofed wicker seats for two people) on the North Sea coast. In month \(m\) (\(m = 1\) for January, …, \(m = 12\) for December), it rents out \(B(m) = 900 - 100(m - 7)^2\) chairs per week during the season from May to September. Next year, the season starts two months earlier. Otherwise, everything stays the same.” Give the function \(B_1(m)\) for next year and name the transformation.

    Check: In which month does next year’s season have its peak? Is it two months before this year’s peak?

  4. An FSP task says: “Give a function \(f_1(x)\) which is a mirror image of \(f(x) = (x - 3)^2 + 1\) mirrored at the \(y\)-axis and give a reason for your choice.” Give \(f_1(x)\) and the reason.

    Check: The vertex of \(f\) is \((3, 1)\). Where must the vertex of the mirror image lie? Does your \(f_1\) have its vertex there?

  5. “The price-demand function of a museum is \(p(x) = 14 - 0.02x\): at a ticket price of \(p(x)\) EUR, \(x\) visitors come per day. The director asks: ‘How many visitors come at a price of 8 EUR?’” Give the function \(x(p)\) for the number of visitors at a price of \(p\) EUR and answer the director’s question.

    Check: Insert your number of visitors into \(p(x)\). Do you get 8 EUR?

  1. The rent is a fixed cost, so every cost value rises by 600 EUR:

    \(C_1(x) = C(x) + 600 = 0.2x^2 + 25x + 5400\), a vertical shift up by 600.

    Check: yes. \(C(0) = 4{,}800\) EUR and \(C_1(0) = 5{,}400\) EUR, a difference of 600 EUR.

    Typical mistake: \(C(x + 600)\). Replacing \(x\) shifts the graph horizontally: \(C(x + 600)\) would be the costs for 600 more participants. A change of the fixed costs shifts the graph vertically.

  2. Every demand value is multiplied by 1.2:

    \(D_1(p) = 1.2 \cdot D(p) = 1.2(800 - 25p) = 960 - 30p\), a vertical stretch by the factor 1.2.

    Check: yes. \(D(32) = 800 - 800 = 0\) and \(D_1(32) = 960 - 960 = 0\). 20 % more of nothing is still nothing: a vertical stretch keeps the zeros.

    Typical mistake: \(D(p) + 20\) (20 bags more, not 20 %) or \(D(1.2p)\). \(D(1.2p) = 800 - 30p\) changes the price instead of the number of bags and would already be 0 at about 26.67 EUR.

  3. Two months earlier: the number of chairs that belonged to month \(m + 2\) now occurs in month \(m\):

    \(B_1(m) = B(m + 2) = 900 - 100(m + 2 - 7)^2 = 900 - 100(m - 5)^2\), a horizontal shift by 2 to the left. The season now runs from March to July.

    Check: yes. This year’s peak is at \(m = 7\) (July), next year’s at \(m = 5\) (May).

    Typical mistake: \(B(m - 2) = 900 - 100(m - 9)^2\) with the peak in September, two months later. Plus shifts left, minus shifts right.

  4. \(f_1(x) = f(-x) = (-x - 3)^2 + 1 = (x + 3)^2 + 1\)

    Reason: replacing \(x\) by \(-x\) maps each point \((a, f(a))\) to \((-a, f(a))\), so the graph is mirrored from right to left at the \(y\)-axis.

    Check: yes. The mirror image of the vertex \((3, 1)\) is \((-3, 1)\), and \((x + 3)^2 + 1\) has its vertex at \((-3, 1)\).

    Typical mistake: \(-f(x) = -(x - 3)^2 - 1\). This is the mirror image at the \(x\)-axis, with the vertex \((3, -1)\) and a parabola that opens downward.

  5. The price-demand function takes a number of visitors and gives a price. The director needs the other direction, the inverse function. Solve \(p = 14 - 0.02x\) for \(x\):

    \(0.02x = 14 - p \iff x(p) = 700 - 50p\)

    At 8 EUR: \(x(8) = 700 - 400 = 300\) visitors per day.

    Check: yes. \(p(300) = 14 - 6 = 8\) EUR.

    Typical mistake: \(p(8) = 14 - 0.16 = 13.84\). The input of \(p(x)\) is a number of visitors, so the result is a price, not a number of visitors.

Problem 6: One Paragraph, All Conditions (xx)

A company prints photo books. For \(x\) photo books per month, its total costs are \(C(x) = 0.01x^2 + 6x + 3500\) EUR, and its customers order according to the price-demand function \(p(x) = 30 - 0.01x\) (price in EUR per book). Next year, the paper supplier raises its prices, so all costs will be 5 % higher. After that, the landlord raises the rent by 300 EUR per month. The owner wants to know the new cost function, the new profit as a function of the price \(p\) (so that they can compare prices directly), and how many photo books the customers order at a price of 22 EUR.

List the new cost function \(C_1(x)\) with its transformations in the correct order, the revenue \(R(x)\), the new profit \(P_1(x)\), the function \(x(p)\) for the number of books at a price \(p\) with its domain, the composite function for the profit as a function of the price, and a formula for the owner’s last question. Do not solve anything.

Check: An assistant writes \(C_1(x) = 1.05 \cdot (C(x) + 300)\). Compare the costs in a month without photo books, \(C_1(0)\), for the assistant’s term and for yours. Which one matches the paragraph?

  1. New costs: \(C_1(x) = 1.05 \cdot C(x) + 300\). First the vertical stretch by 1.05 (all costs 5 % higher), then the vertical shift up by 300 (the rent increase comes afterwards and is not raised by 5 %)
  2. Revenue: \(R(x) = x \cdot p(x) = 30x - 0.01x^2\)
  3. New profit: \(P_1(x) = R(x) - C_1(x)\)
  4. Number of books at a price \(p\): the inverse of the price-demand function. Solve \(p = 30 - 0.01x\) for \(x\): \(x(p) = 3000 - 100p\) with \(0 \leq p \leq 30\)
  5. Profit as a function of the price: \((P_1 \circ x)(p) = P_1(x(p))\). First \(x(p)\) gives the number of books, then \(P_1\) gives the profit for this number
  6. Books at 22 EUR: \(x(22)\)

Check: with your term, \(C_1(0) = 1.05 \cdot 3500 + 300 = 3{,}975\) EUR. With the assistant’s term, \(1.05 \cdot (3500 + 300) = 3{,}990\) EUR. The paragraph raises all costs by 5 % first (\(3{,}675\) EUR) and then adds 300 EUR, which gives 3,975 EUR. The assistant’s term also raises the rent increase by 5 %.

Typical mistake: \((x \circ P_1)(p)\), the wrong order. \(P_1\) needs a number of books as its input, and \(x(p)\) delivers it. So \(x(p)\) is inserted into \(P_1\), not the other way round.

C: Applied

Problem 7: Costs of a Hammock Manufactory (xx)

A small company sews hammocks. Its monthly costs for \(x\) hammocks are

\[C(x) = 0.02x^2 + 15x + 2000 \text{ EUR}\]

The term \(0.02x^2\) describes extra costs at high outputs (overtime), \(15x\) the material per hammock, and 2000 the fixed costs (rent, machines, base salaries).

  1. A new fabric supplier lowers the variable costs (the two terms that depend on \(x\)) by 20 %. Give the new cost function \(C_1(x)\). Explain why \(C_1\) is not a vertical compression of \(C\).

  2. The company moves into a larger workshop, so its fixed costs rise by 1,500 EUR per month. Give the cost function \(C_2(x)\) for this change alone and name the transformation.

  3. Give the cost function \(C_3(x)\) with both changes and decide whether the order of the two changes matters. Compare with a case in which all costs fall by 20 %: give the terms for “first 20 % less, then 1,500 EUR more” and for “first 1,500 EUR more, then 20 % less”.

(for verification: \(C_3(x) = 0.016x^2 + 12x + 3500\))

  1. Compute the monthly costs for 100 hammocks before and after both changes. Decide which is cheaper and explain why.

  2. Determine the output at which the costs before and after both changes are equal, rounded to two decimal places. Decide from which number of hammocks per month on the new situation is cheaper.

(for verification: about 343 hammocks)

a) Only the two variable terms are multiplied by 0.8:

\[C_1(x) = 0.8 \cdot 0.02x^2 + 0.8 \cdot 15x + 2000 = 0.016x^2 + 12x + 2000\]

A vertical compression by 0.8 would multiply all of \(C\), including the fixed costs: \(0.8 \cdot C(x) = 0.016x^2 + 12x + 1600\). But the fixed costs stay at 2,000 EUR: \(C_1(0) = 2000\), not \(0.8 \cdot C(0) = 1600\).

b) \(C_2(x) = C(x) + 1500 = 0.02x^2 + 15x + 3500\), a vertical shift up by 1,500.

c) \(C_3(x) = 0.016x^2 + 12x + 3500\). Here the order does not matter: the 20 % only change the terms with \(x\), and the 1,500 EUR only change the constant. Both orders give the same function.

If all costs fell by 20 %, the order would matter:

  • First 20 % less, then 1,500 EUR more: \(0.8 \cdot C(x) + 1500 = 0.016x^2 + 12x + 3100\)
  • First 1,500 EUR more, then 20 % less: \(0.8 \cdot (C(x) + 1500) = 0.016x^2 + 12x + 2800\)

A compression after the shift also compresses the shift (1,500 EUR become 1,200 EUR).

d) \(C(100) = 200 + 1500 + 2000 = 3{,}700\) EUR and \(C_3(100) = 160 + 1200 + 3500 = 4{,}860\) EUR.

Before the changes, 100 hammocks are 1,160 EUR cheaper. The variable costs fall by only \(40 + 300 = 340\) EUR, while the fixed costs rise by 1,500 EUR.

e) Condition \(C(x) = C_3(x)\):

\[0.02x^2 + 15x + 2000 = 0.016x^2 + 12x + 3500 \iff 0.004x^2 + 3x - 1500 = 0 \;\big|\cdot 250 \iff x^2 + 750x - 375000 = 0\]

Quadratic formula with \(\Delta = 750^2 + 4 \cdot 375000 = 562500 + 1500000 = 2062500\) and \(\sqrt{2062500} \approx 1436.14\):

\[x_{1,2} = \frac{-750 \pm 1436.14}{2}, \qquad x_1 \approx 343.07, \qquad x_2 \approx -1093.07\]

\(x_2\) is negative and no output. At about 343.07 hammocks per month, both cost functions are equal.

The difference \(C(x) - C_3(x) = 0.004x^2 + 3x - 1500\) is an upward parabola, negative between its zeros. So below 343.07 hammocks the old costs are lower, above it the new ones. Test value: \(C(400) = 3200 + 6000 + 2000 = 11{,}200\) EUR and \(C_3(400) = 2560 + 4800 + 3500 = 10{,}860\) EUR ✓

The new situation is cheaper from 344 hammocks per month on. At lower outputs, the higher fixed costs outweigh the savings.

Problem 8: Subscriptions of a Hiking App (xx)

A start-up sells yearly subscriptions to its hiking app (an app with hiking routes and maps). If it sells \(x\) thousand subscriptions per year, its revenue is

\[R(x) = -2x^2 + 120x \text{ thousand EUR}\]

The price per subscription is then \(120 - 2x\) EUR: the more subscriptions the start-up wants to sell, the lower the price must be.

  1. Compute the number of subscriptions with the highest revenue and prove that it is a maximum. Give the maximum revenue and the price per subscription.

  2. With a new offline map, the customers pay 25 % more at every number of subscriptions, so the revenue is \(R_1(x) = 1.25 \cdot R(x)\). Name the transformation and give \(R_1\) in standard form. Show that the number of subscriptions with the highest revenue stays the same and give the new maximum revenue.

(for verification: 2,250 thousand EUR)

  1. The start-up’s costs are \(C(x) = 30x + 600\) thousand EUR per year (servers and map licences per subscription, and the development team). Give the profit functions \(P(x)\) before and \(P_1(x)\) after the offline map. Compute the profit-maximizing number of subscriptions and the maximum profit in both cases. Explain why the profit-maximizing number changes, although the revenue maximum stays at 30,000 subscriptions.

(for verification: 22,500 and 24,000 subscriptions)

  1. Now a competitor launches a free hiking app. Instead of the offline map scenario, the start-up expects 25 % less revenue than \(R(x)\) at every number of subscriptions. Give the new revenue \(R_2(x)\) and show that the profit function is then \(P_2(x) = -1.5(x - 20)^2\). Interpret the result.

  2. The state’s tourism office offers the start-up a fixed payment of 150 thousand EUR per year if the app shows the office’s hiking routes. Describe the effect on \(P_2\) as a transformation. Decide whether the best number of subscriptions changes and give the maximum profit.

a) Vertex formula: \(x_v = -\frac{120}{2 \cdot (-2)} = 30\). Since \(a = -2 < 0\), the parabola opens downward, so the vertex is a maximum.

  • Number of subscriptions: 30 thousand \(= 30{,}000\) per year
  • Maximum revenue: \(R(30) = -1800 + 3600 = 1{,}800\) thousand EUR \(= 1.8\) million EUR
  • Price: \(120 - 2 \cdot 30 = 60\) EUR per subscription

b) Every revenue value is multiplied by 1.25: a vertical stretch by the factor 1.25.

\[R_1(x) = 1.25 \cdot (-2x^2 + 120x) = -2.5x^2 + 150x\]

Vertex formula: \(x_v = -\frac{150}{2 \cdot (-2.5)} = 30\), the same as before, and a maximum since \(a = -2.5 < 0\). The factor 1.25 multiplies \(a\) and \(b\), so \(-\frac{b}{2a}\) does not change: the graph is stretched upward, and the vertex stays above \(x = 30\).

New maximum revenue: \(R_1(30) = -2250 + 4500 = 2{,}250\) thousand EUR \(= 1.25 \cdot 1800\) ✓

c) Profit = revenue − costs:

\[P(x) = -2x^2 + 120x - (30x + 600) = -2x^2 + 90x - 600\]

\[P_1(x) = -2.5x^2 + 150x - (30x + 600) = -2.5x^2 + 120x - 600\]

  • Before: \(x_v = -\frac{90}{2 \cdot (-2)} = 22.5\), i.e. 22,500 subscriptions, \(P(22.5) = -1012.5 + 2025 - 600 = 412.5\) thousand EUR \(= 412{,}500\) EUR
  • After: \(x_v = -\frac{120}{2 \cdot (-2.5)} = 24\), i.e. 24,000 subscriptions, \(P_1(24) = -1440 + 2880 - 600 = 840\) thousand EUR \(= 840{,}000\) EUR

Both are maxima, since \(a < 0\).

Only the revenue is stretched, the costs stay the same. So \(P_1 = 1.25 \cdot R - C\) is not \(1.25 \cdot P\), and the graph of \(P\) is not just stretched: \(a\) changes from \(-2\) to \(-2.5\), but \(b\) from 90 to 120. Every subscription now brings more revenue for the same costs, so it pays to sell more.

d) \(R_2(x) = 0.75 \cdot R(x) = -1.5x^2 + 90x\) (a vertical compression by 0.75).

\[P_2(x) = -1.5x^2 + 90x - (30x + 600) = -1.5x^2 + 60x - 600 = -1.5(x^2 - 40x + 400) = -1.5(x - 20)^2 \quad \checkmark\]

The bracket is a perfect square: \((x - 20)^2 = x^2 - 40x + 400\).

A square is never negative, so \(P_2(x) \leq 0\) for every \(x\). The start-up can at best break even, with 20,000 subscriptions (\(P_2(20) = 0\)). Every other number of subscriptions means a loss.

e) \(P_3(x) = P_2(x) + 150 = -1.5(x - 20)^2 + 150\), a vertical shift up by 150.

The vertex moves from \((20, 0)\) to \((20, 150)\). The best number of subscriptions stays 20,000, now with a maximum profit of 150 thousand EUR \(= 150{,}000\) EUR per year. A vertical shift moves the graph up, but not to the left or right.

Problem 9: Production Costs of a Perfume Manufactory (xx)

A small perfume manufactory in Hamburg makes one batch of a perfume in three stages:

  • Stage 1 (mixing): \(x\) litres of raw materials (essential oils and alcohol) are mixed. Costs: \(C_1(x) = 30x + 150\) EUR.
  • Stage 2 (maturing and filtering): 1 litre of the mixture gives 0.7 litres of perfume, the rest stays in the filter. Costs: \(C_2(y) = 50y + 180\) EUR for \(y\) litres of perfume.
  • Stage 3 (filling): each litre of perfume fills 5 bottles of 200 ml. Costs for bottles and packaging: \(C_3(z) = 2z + 100\) EUR for \(z\) bottles.

  1. Give the litres of perfume \(y(x)\) and the number of bottles \(z(y)\) as functions. Give a term for the composite function \((z \circ y)(x)\) and explain its meaning.

  2. Show that the total costs of a batch are \(C(x) = 72x + 430\) EUR. Explain which of the three cost terms are composite functions.

  3. A perfume shop orders 100 bottles. Determine the inverse function of \((z \circ y)(x)\) and explain its meaning. Compute how many litres of raw materials are needed and the total costs. The raw materials are only sold in whole litres: decide how many litres the manufactory buys.

  4. Give a term for the total costs as a function of the number of bottles, \(K(z) = C(x(z))\), where \(x(z)\) is your inverse function from c). Explain why you need the inverse function for this. Compute the average cost per bottle for a batch of 35 bottles.

(for verification: 1,150 EUR for 35 bottles)

  1. The manufactory sells each bottle to perfume shops for 49 EUR. Determine from which number of bottles per batch on it makes a profit.

a) \(y(x) = 0.7x\) litres of perfume and \(z(y) = 5y\) bottles.

\[(z \circ y)(x) = z(y(x)) = 5 \cdot 0.7x = 3.5x\]

It gives the number of bottles that \(x\) litres of raw materials fill: first the filter step (\(y\)), then the filling (\(z\)).

b) Every stage has its own input, so the costs of stages 2 and 3 have to be written with \(x\):

\[C(x) = C_1(x) + C_2(y(x)) + C_3\big((z \circ y)(x)\big) = (30x + 150) + (50 \cdot 0.7x + 180) + (2 \cdot 3.5x + 100)\]

\[C(x) = 30x + 150 + 35x + 180 + 7x + 100 = 72x + 430 \quad \checkmark\]

\(C_2(y(x))\) and \(C_3((z \circ y)(x))\) are composite functions: the costs of stage 2 depend on the litres of perfume, which depend on \(x\). The costs of stage 3 depend on the bottles, which depend on the litres of perfume and therefore on \(x\).

c) Solve \(z = 3.5x\) for \(x\): \(x(z) = \frac{z}{3.5}\). The inverse function gives the litres of raw materials needed for \(z\) bottles.

For 100 bottles: \(x(100) = \frac{100}{3.5} \approx 28.57\) litres. Total costs: \(C\left(\frac{100}{3.5}\right) = 72 \cdot \frac{100}{3.5} + 430 = \frac{14400}{7} + 430 \approx 2057.14 + 430 = 2{,}487.14\) EUR.

Whole litres: 28 litres give only \(3.5 \cdot 28 = 98\) bottles, which is too few. The manufactory buys 29 litres (enough for 101 bottles).

d) Insert the inverse function into \(C\):

\[K(z) = C(x(z)) = 72 \cdot \frac{z}{3.5} + 430 = \frac{144}{7}z + 430 \approx 20.57z + 430\]

\(C\) needs litres of raw materials as its input, but the order is given in bottles. The inverse function translates bottles into litres, then \(C\) gives the costs.

For 35 bottles: \(K(35) = \frac{144}{7} \cdot 35 + 430 = 720 + 430 = 1{,}150\) EUR (this is \(C(10)\), a batch from 10 litres). Average cost: \(\frac{1150}{35} \approx 32.86\) EUR per bottle.

e) Profit when the revenue exceeds the costs:

\[49z > \frac{144}{7}z + 430 \iff \frac{343 - 144}{7}z > 430 \iff \frac{199}{7}z > 430 \iff z > \frac{3010}{199} \approx 15.13\]

The manufactory makes a profit from 16 bottles per batch on. Check: \(K(16) \approx 759.14\) EUR \(< 49 \cdot 16 = 784\) EUR ✓, while \(K(15) \approx 738.57\) EUR \(> 49 \cdot 15 = 735\) EUR.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Inflatable Stand-Up Paddle Boards [21 pts. total]

A company develops inflatable stand-up paddle boards (SUP boards: boards on which you stand and paddle across a lake) and sells them to sports shops. For one season, its price-demand function and its total costs are

\[p(x) = 90 - 2.5x, \qquad C(x) = 0.5x^2 + 30x + 150\]

with \(p\) in MU per QU, \(C\) in MU and \(x\) in QU per season. The term \(0.5x^2\) describes costs per board that rise with the output, for example for overtime.

(QU = quantity units, MU = monetary units. 1 QU = 100 boards, 1 MU = 1,000 EUR, so a price of \(p\) MU per QU is a price of \(10p\) EUR per board. Revenue = quantity × price. Profit = revenue − total costs.)

Part a)

Give the revenue function \(R(x)\). Show that the profit function is given by \(P(x) = -3x^2 + 60x - 150\). [2 pts.]

Revenue = quantity × price:

\[R(x) = x \cdot p(x) = x(90 - 2.5x) = 90x - 2.5x^2\]

Profit = revenue − total costs. The minus sign changes every sign in the second bracket:

\[P(x) = 90x - 2.5x^2 - (0.5x^2 + 30x + 150) = -3x^2 + 60x - 150 \quad \checkmark\]

Grading: 1 Pt. \(R(x)\) and ansatz \(P = R - C\), 1 Pt. sign change and collecting terms.

Part b)

Determine the (lower) break-even point and the profit limit (upper break-even point) in QU, rounded to two decimal places. Decide how many boards per season the company must sell to make a profit. [4 pts.]

(for verification: break-even point at about 2.93 QU)

Zero form. At a break-even point the profit is zero, \(P(x) = 0\). Divide by \(-3\):

\[-3x^2 + 60x - 150 = 0 \;\big|: (-3) \iff x^2 - 20x + 50 = 0\]

Quadratic formula with \(a = 1\), \(b = -20\), \(c = 50\): \(\Delta = (-20)^2 - 4 \cdot 1 \cdot 50 = 400 - 200 = 200\) and \(\sqrt{200} = 10\sqrt{2}\):

\[x_{1,2} = \frac{20 \pm 10\sqrt{2}}{2} = 10 \pm 5\sqrt{2}, \qquad x_1 \approx 2.93, \qquad x_2 \approx 17.07\]

Break-even point: \(x_1 \approx 2.93\) QU. Profit limit: \(x_2 \approx 17.07\) QU.

Profit zone. The parabola opens downward (\(a = -3 < 0\)), so the profit is positive between the zeros. Test value: \(P(10) = -300 + 600 - 150 = 150 > 0\) ✓

In boards: \(100 \cdot (10 - 5\sqrt{2}) \approx 292.9\) and \(100 \cdot (10 + 5\sqrt{2}) \approx 1707.1\). Only whole boards are sold, so the company makes a profit with 293 to 1,707 boards per season (both numbers lie inside the zeros).

Grading: 1 Pt. zero form, 1 Pt. quadratic formula with discriminant, 1 Pt. break-even point and profit limit named, 1 Pt. whole boards 293 to 1,707 with reason (\(a < 0\)).

Part c)

The sports shops sell the boards to end customers. A shop adds 30 % to the company’s price \(q\) (in EUR per board) and 20 EUR for a repair kit: \(M(q) = 1.3q + 20\). On this price, the customer pays 19 % VAT (value added tax): \(V(r) = 1.19r\), where \(r\) is the shop’s price without VAT.

Give a term for the composite function \((V \circ M)(q)\) and explain its meaning. Explain why \((M \circ V)(q)\) does not describe the price the customer pays. A shop sells a board for 952 EUR. Calculate the company’s price per board. A trainee says: “The customer pays 30 % + 19 % = 49 % more than the company’s price.” Check this claim for this board. [5 pts.]

(for verification: 600 EUR per board)

Composite function: first \(M\), then \(V\):

\[(V \circ M)(q) = V(M(q)) = 1.19(1.3q + 20) = 1.547q + 23.8\]

It gives the price in EUR (with VAT) that the end customer pays in the shop if the company asks \(q\) EUR per board.

Order. \((M \circ V)(q) = 1.3 \cdot 1.19q + 20 = 1.547q + 20\) would add the VAT before the shop’s markup, and the 20 EUR for the repair kit would carry no VAT. The VAT is charged on the shop’s whole price, so \(V\) is the outer function and comes last.

Company’s price:

\[1.19(1.3q + 20) = 952 \iff 1.3q + 20 = 800 \iff 1.3q = 780 \iff q = 600 \text{ EUR per board}\]

Trainee’s claim: \(\frac{952 - 600}{600} \approx 0.5867\), so the customer pays about 58.67 % more, not 49 %. The two percentages multiply (\(1.3 \cdot 1.19 = 1.547\), i.e. 54.7 %), and the repair kit comes on top.

Grading: 1 Pt. composite term, 1 Pt. meaning, 1 Pt. order with reason, 1 Pt. company’s price 600 EUR, 1 Pt. 58.67 % with reason.

Part d)

Determine the inverse function \(x(p)\) of the price-demand function and explain its meaning. The company asks for a price of 600 EUR per board. Calculate the number of boards sold and the profit in EUR. [5 pts.]

Inverse function. Solve \(p = 90 - 2.5x\) for \(x\):

\[2.5x = 90 - p \iff x(p) = 36 - 0.4p, \qquad 0 \leq p \leq 90\]

It gives the number of QU that the shops buy per season if the company asks \(p\) MU per QU. Check: \(x(90) = 0\), at the maximum price of 900 EUR per board nobody buys.

Units. 600 EUR per board \(= 600 \cdot 100 = 60{,}000\) EUR per QU \(= 60\) MU per QU.

Boards sold: \(x(60) = 36 - 24 = 12\) QU \(= 1{,}200\) boards per season. Check: \(p(12) = 90 - 30 = 60\) ✓

Profit:

\[P(12) = -3 \cdot 144 + 60 \cdot 12 - 150 = -432 + 720 - 150 = 138 \text{ MU} = 138{,}000 \text{ EUR}\]

1,200 boards lie in the profit zone from b).

Grading: 1 Pt. inverse function, 1 Pt. meaning, 1 Pt. conversion to 60 MU per QU, 1 Pt. 1,200 boards, 1 Pt. profit of 138,000 EUR.

Part e)

Besides the shops, the company sells boards in its own online shop. In month \(m\) (\(m = 1\) for January, …, \(m = 12\) for December), it receives \(s(m) = 150 - 6(m - 8)^2\) online orders during the season from April to December. For the next season, the company expects three changes: the season starts one month earlier, a TV report makes SUP boards more popular so that there are 20 % more orders in every month, and a sports club orders 30 additional boards every month through the online shop.

Give a function \(s_2(m)\) which describes the online orders per month in the next season and give a reason for your choice. Name the transformations in the correct order. Determine the month with the most online orders and their number. [5 pts.]

(for verification: \(s_2(m) = 210 - 7.2(m - 7)^2\))

\[s_2(m) = 1.2 \cdot s(m + 1) + 30\]

Reason:

  1. One month earlier: the orders that came in month \(m + 1\) now come in month \(m\), so \(m\) is replaced by \(m + 1\). This is a horizontal shift by 1 to the left
  2. 20 % more orders in every month: all values are multiplied by 1.2, a vertical stretch
  3. The club’s 30 boards per month are added last, a vertical shift up by 30. They are not part of the 20 % increase: \(1.2 \cdot (s(m + 1) + 30) = 1.2 \cdot s(m + 1) + 36\) would raise them to 36

Vertex form:

\[s_2(m) = 1.2\left(150 - 6(m + 1 - 8)^2\right) + 30 = 180 - 7.2(m - 7)^2 + 30 = 210 - 7.2(m - 7)^2\]

Vertex \((7, 210)\). Since \(-7.2 < 0\), the parabola opens downward, so the vertex is a maximum. The most online orders come in July (\(m = 7\)), with 210 boards, one month earlier and 60 boards more than this season (August, \(s(8) = 150\)). Check: \(s_2(7) = 1.2 \cdot s(8) + 30 = 180 + 30 = 210\) ✓

Grading: 1 Pt. \(s(m + 1)\) with reason, 1 Pt. factor 1.2 before \(+ 30\) with reason, 1 Pt. the three transformations in the correct order, 1 Pt. vertex form, 1 Pt. July with 210 boards.

Calculator Practice

Problem 10: Calculator: Composite Functions in Table Mode (x)

Press MENU → 9 (Tabellen) and enter \(f(x) = 3x - 7\) and \(g(x) = x^2 - 4x + 2\). Press = and set Start \(= -4\), Ende \(= 4\) and Inkre \(= 1\).

In a) to d), compute the value with the table: for \(f(g(3))\), read \(g(3)\) in the \(g\) column, then look up the value of \(f\) at this number in the \(f\) column.

  1. \(f(g(3))\)

  2. \(g(f(3))\)

  3. \(f(g(0))\)

  4. \(g(f(1))\)

  5. Explain with a) and b) why \(f \circ g\) and \(g \circ f\) are different functions. Check a) by hand with the term of \((f \circ g)(x)\).

  6. Decide whether your table is enough to compute \(g(f(4))\). If not, give a new value for Ende and the result.

The table for \(x = -4, -3, \ldots, 4\):

\(x\) \(-4\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(f(x)\) \(-19\) \(-16\) \(-13\) \(-10\) \(-7\) \(-4\) \(-1\) \(2\) \(5\)
\(g(x)\) \(34\) \(23\) \(14\) \(7\) \(2\) \(-1\) \(-2\) \(-1\) \(2\)

  1. \(g(3) = -1\) (column \(x = 3\)), then \(f(-1) = -10\) (column \(x = -1\)). So \(f(g(3)) = -10\)

  2. \(f(3) = 2\), then \(g(2) = -2\). So \(g(f(3)) = -2\)

  3. \(g(0) = 2\), then \(f(2) = -1\). So \(f(g(0)) = -1\)

  4. \(f(1) = -4\), then \(g(-4) = 34\). So \(g(f(1)) = 34\)

  5. For the same input 3, the two orders give \(-10\) and \(-2\). So \(f \circ g\) and \(g \circ f\) are different functions: the order of a composition matters.

    By hand: \((f \circ g)(x) = 3(x^2 - 4x + 2) - 7 = 3x^2 - 12x - 1\) and \((f \circ g)(3) = 27 - 36 - 1 = -10\) ✓

  6. \(f(4) = 5\), but the table ends at \(x = 4\), so \(g(5)\) is missing. With Ende \(= 5\), the table shows \(g(5) = 25 - 20 + 2 = 7\). So \(g(f(4)) = 7\). The table must contain every inner value as an \(x\)-value.